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scoundrel [369]
3 years ago
11

Julia is taking a test with multiple choice and true/false question the test has twentg question altogether . julia gets five po

ints for correct multiple choice answer nd lose one points for each incorrect true/fales answers . she earned 76 points in the test how many multilpe choice question and true/fales question are one the test ? .
let ___ =________
let ___=________
system of equestion :




solution :

plzz help fastt :)
Mathematics
1 answer:
Sonbull [250]3 years ago
5 0
X=number of corect
y=number of incorrect
5 points for each correct so 5x
-1 point for each incorrect so -y

5x-y=76
solve for x and y

I like putting it into slope intercep tofrm which is y=mx+b
add y to both sides
5x=76+y
subtract 76 from both sides
5x-75=y
y=5x-76

subsitute values for x and get values for y
remember, that julia cannot answer a negative number of incorrect questions so y must be greater than or equal to 0

so solve for minimum
0<u><</u>5x-76
add 76 to both sides
76<u><</u>5x
divide by 5
15.2<u><</u>x
you cannot answer a fraction of a correct quesiton so round up
16 is the minumum number of queswtions correctlyh answered


so
x<u>></u>16

subsitute 16 for x
y=5(16)-76
y=80-76
y=4

as we subsitute more numbers over 16, we get these points
so (x,y) 
(16,4)
(17,9)
(18,14)
(19,19)
(20,24)
(21,29)
(22,34)
(23,39)
x=correct answer
y=wrong answers
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3 years ago
A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0
yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

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Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

P(x

Where, for example, P(0) and P(1) are calculated as:

P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098

Then, P(x, so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)

Where, for example, P(4) is calculated as:

P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001

Then, P(x\geq4 )=0.3504, so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0
3 years ago
8. Hassan made a vegetable salad with 2 3/8 pounds
amm1812

let's convert firstly the mixed fractions to improper fractions and then add up.

\bf \stackrel{mixed}{2\frac{3}{8}}\implies \cfrac{2\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{19}{8}}~\hfill \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}} \\\\\\ \stackrel{mixed}{2\frac{7}{8}}\implies \cfrac{2\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{23}{8}} \\\\[-0.35em] ~\dotfill

\bf \cfrac{19}{8}+\cfrac{5}{4}+\cfrac{23}{8}\implies \stackrel{\textit{using an LCD of 8}}{\implies \cfrac{(1)19+(2)5+(1)23}{8}}\implies \cfrac{19+10+23}{8} \\\\\\ \cfrac{52}{8}\implies \cfrac{13}{2}\implies 6\frac{1}{2}

5 0
3 years ago
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